It appears you are looking for sufficient conditions in terms of expectations of functions of $X_t$. If so, the following may be offered:

\begin{equation}
\text{$|X_t|$ diverges to $\infty$ in probability iff $E\frac1{1+|X_t|}\to0$;} \tag{*}
\end{equation}
here everywhere the convergence is as $t\to\infty$.

Indeed, $|X_t|$ diverges to $\infty$ in probability iff $Y_t:=1/|X_t|$ converges to $0$ in probability, that is, iff $P(Y_t>b)\to0$ for each real $b>0$. Let $f(y):=\frac y{1+y}$.

Then $I\{y>b\}\le f(y)/f(b)$ for $b>0$ and $y\ge0$, where $I\{\cdot\}$ is the indicator function. So, $P(Y_t>b)\le Ef(Y_t)/f(b)=E\frac1{1+|X_t|}/f(b)\to0$ if $E\frac1{1+|X_t|}\to0$, which proves the "if" part of (*). (In fact, here we use the Markov inequality.)

Using inequality $f(y)\le I\{y>b\}+f(b)$ for $b>0$ and $y\ge0$, we see that
\begin{equation}
E\frac1{1+|X_t|}= Ef(Y_t)\le P(Y_t>b) +f(b)\to f(b)
\end{equation}
for each real $b>0$ if $Y_t$ converges to $0$ in probability. So, the "only if" part of (*) follows as well, by letting $b\downarrow0$. (This part also follows by the Lebesgue dominated convergence theorem.)

It should also be clear that, in (*), one can replace $E\frac1{1+|X_t|}$ by $E\frac1{1+X_t^2}$ or $E\exp(-X_t^2/2)$ or, more generally, by $Eg(|X_t|)$, where $g$ is (say) any continuous function which is strictly decreasing to $0$ on $[0,\infty)$.

In particular, this allows one to express the condition that $|X_t|$ diverges to $\infty$ in terms of the characteristic functions $c_t(u)=Ee^{iuX_t}$ of $X_t$ -- say, by writing $e^{-x^2/2}=\int_{-\infty}^\infty e^{iux}\frac1{\sqrt{2\pi}}e^{-u^2/2}\,du$ and $\frac2{1+x^2}=\int_{-\infty}^\infty e^{iux}e^{-|u|}\,du$ for real $x$, whence
$E\exp(-X_t^2/2)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty c_t(u)e^{-u^2/2}\,du$ and
$E\frac1{1+X_t^2}=\frac12\,\int_{-\infty}^\infty c_t(u)e^{-|u|}\,du$, and so,
\begin{equation}
\text{$|X_t|\to\infty$ in probability iff $\int_{-\infty}^\infty c_t(u)e^{-u^2/2}\,du\to0$ iff $\int_{-\infty}^\infty c_t(u)e^{-|u|}\,du\to0$.}
\end{equation}